AoPS - Problem

Source: RMM Extralist 2021 A3

Tags: RMM Shortlist, algebra, intervals, covering, combinatorics

Assassino9931

18.09.2023 23:49

A tile $T$ is a union of finitely many pairwise disjoint arcs of a unit circle $K$. The size of $T$, denoted by $|T|$, is the sum of the lengths of the arcs $T$ consists of, divided by $2\pi$. A copy of $T$ is a tile $T'$ obtained by rotating $T$ about the centre of $K$ through some angle. Given a positive real number $\varepsilon < 1$, does there exist an infinite sequence of tiles $T_1,T_2,\ldots,T_n,\ldots$ satisfying the following two conditions simultaneously: 1) $|T_n| > 1 - \varepsilon$ for all $n$; 2) The union of all $T_n'$ (as $n$ runs through the positive integers) is a proper subset of $K$ for any choice of the copies $T_1'$, $T_2'$, $\ldots$, $T_n', \ldots$?

HIDE: Note In the extralist the problem statement had the clause "three conditions" rather than two, but only two are presented, the ones you see. I am quite confident this is a typo or that the problem might have been reformulated after submission.

Assassino9931 wrote: ... 2) The union of all $T_n'$ (as $n$ runs through the positive integers) is not a subset of $K$ for any choice of the copies $T_1'$, $T_2'$, $\ldots$, $T_n', \ldots$? I suppose the red text should be read as "is a proper subset of " $K$. Otherwise it's meaningless. Such sequence does exist, no matter how small $\varepsilon>0$ is. Let $U_n$ be the complement of $T_n$. It boils down to construct a sequence $U_n, n\in\mathbb{N}, |U_n|<\varepsilon$, such that $\cap_{n=1}^{\infty}U'_n \neq \emptyset$. This could be done if $T_n$ consists of many small intervals of length $\delta_n>0$ spaced with much bigger intervals of length $\delta'_n>0$ where $\delta/(\delta+\delta'))<\varepsilon$. Choosing $\delta_n$ as rapidly decreasing sequence, we can guarantee that there is a nested sequence of closed intervals $\Delta_n, \Delta_n\subset U_n$, no matter how $U_n$ is rotated. Since $\cap_{n=1}^{\infty}\Delta_n\neq \emptyset$, the result follows.